Cube Stacking

题目描述

Language:
Cube Stacking
Time Limit: 2000MS Memory Limit: 30000K
Total Submissions: 30051 Accepted: 10556
Case Time Limit: 1000MS
Description

Farmer John and Betsy are playing a game with N (1 <= N <= 30,000)identical cubes labeled 1 through N. They start with N stacks, each containing a single cube. Farmer John asks Betsy to perform P (1<= P <= 100,000) operation. There are two types of operations:
moves and counts.

In a move operation, Farmer John asks Bessie to move the stack containing cube X on top of the stack containing cube Y.
In a count operation, Farmer John asks Bessie to count the number of cubes on the stack with cube X that are under the cube X and report that value.

Write a program that can verify the results of the game.
Input

Line 1: A single integer, P
Lines 2…P+1: Each of these lines describes a legal operation. Line 2 describes the first operation, etc. Each line begins with a ‘M’ for a move operation or a ‘C’ for a count operation. For move operations, the line also contains two integers: X and Y.For count operations, the line also contains a single integer: X.

Note that the value for N does not appear in the input file. No move operation will request a move a stack onto itself.
Output

Print the output from each of the count operations in the same order as the input file.
Sample Input

6
M 1 6
C 1
M 2 4
M 2 6
C 3
C 4
Sample Output

1
0
2
Source

USACO 2004 U S Open

我先吐槽一波嗯嗯……难受 mmp 了递归实现并查集真是无敌了这种题题意看的朦朦胧胧的我真是弱爆了看别人的题解都得好一会吭哧

真心题意很难懂就是一开始一堆就一个盒子完了有两种（秀）操作，M A B 把A盒子所在的这一坨放到B所在的这一坨上面，注意是以坨为单位放的，放完之后就变成了一坨 C A就是问咱A盒子所在的这一坨中（其实就是一坨）A下面还有几个盒子，这都神吗鬼操作……代码奉上如果一时半会看的听懵逼没关系懵逼完之后才会出奇迹

#include<cstdio>
#include<iostream>//唉……并查集加上递归简直无敌了 我虚了虚了 看题看的都朦朦胧胧的 

using namespace std;
const int maxn=3*1e4+7;

int N,P;
int pre[maxn];//存放i最近一次操作的堆顶节点
int dis[maxn];//i到最近堆顶的距离，是不断更新的因为堆顶一直在变
int rank[maxn];//i所在的堆的高度，也是不断更新的以为堆顶一直在变
char s[3];
int Find(int x)
{
	if(x!=pre[x])//广大码友们注意了哈 这儿是if 不是while 大爷的 第一次看别人题解都瞧走眼了 找了好半天bug
	{
		int t=pre[x];
		pre[x]=Find(t);//一层一层的更新dis dis dis aaaaaa
		dis[x]+=dis[t];//好一个堆的叠加，当前节点到堆顶的距离加上次堆顶到上一堆顶的距离，递归结束后dis[x]存的就是到最高堆顶的距离 
	}
	return pre[x];//返回的也是x的最高堆顶节点 
}
void John(int x,int y)
{
	int fx=Find(x);
	int fy=Find(y);
	if(fx!=fy)
	{
		pre[fy]=fx;//fx变成了新的堆顶
		dis[fy]=rank[fx];//fy到fx所在的这个堆堆顶的距离 
		rank[fx]+=rank[fy];//更新fx这个堆的高度 
	}
	return ;
}
int main()
{
	int A,B;
	while(~scanf("%d",&P))
	{
		int i,j,k;
		for(i=1;i<=30000;i++)
		{
			pre[i]=i;
			rank[i]=1;
			dis[i]=0;
		} 
		for(i=1;i<=P;i++)
		{
			scanf("%s",s);
			if(s[0]=='M')
			{
				scanf("%d%d",&A,&B);
				John(A,B);
			}
			else
			{
				scanf("%d",&A);
				int t=Find(A);
				printf("%d\n",rank[t]-dis[A]-1);//堆的高度减去A到堆顶的距离就是下面hao
			}
		}
	}
	return 0;
}